## Qual Attempt #4

*Date Written: August 30, 2023; Last Modified: August 30, 2023*

I did attempt two more quals before this one, one in algebra and one in analysis. The algebra qual was from 15 years ago, and hence the curriculum (and difficulty) was rather dated. The other analysis qual was difficult, but solutions already exist online, and I had to reference them extensively.

The qual I took today (and, to some extent, last night) was the analysis qual from fall 2022, and my friend Elias helped me with a lot of the problems. Together, we were able to produce solutions for most of the problems, and seeing that solutions weren’t readily available online, I thought I’d write them up and add commentary here.

I did attempt this qual by hand, but the work there is far from complete. Elias and I determined the solutions to several of the missing problems long after the fact.

I’m still working on the solutions, so I’ll be filling in the missing ones over the coming days or weeks.

**Problem 1. **

Let \(f\in L^1\left(\mathbb R^d\right)\) and \(a\in\mathbb R^d\setminus\lbrace 0\rbrace\). For each \(k\in\mathbb N\), determine the limit \[\lim_{t\to\infty}\int \left\lvert \sum_{j=1}^k f(jx+ta)\right\rvert dx.\]

One of the challenges of this problem is that the absolute value signs are outside the summation, and so if \(f\) oscillates unpredictably, there could be some cancellation within the sum that is difficult to estimate. However, since \(f\in L^1\), most of the “action” in the integral occurs on some compact set. Hence, when \(t\) is very, very large, \(jx+at\) being near the “action” for one \(j\) makes the contribution from the rest of the sum very small. I think it helps to draw a picture when \(f\) is a characteristic function.

Based on this intuition, it seems that for \(t\) sufficiently large, we would get \[\left\lvert\sum_{j=1}^k f(jx+at)\right\rvert \approx \sum_{j=1}^k \left\lvert f(jx+at)\right\rvert.\] In fact, when \(f\) has comact support, this seems to become an equality when \(t\) is very large.

To formalise this, fix \(\epsilon>0\). We let \(g\in C_c^\infty\cap L^1\) such that \(\left\lVert f-g\right\rVert<\epsilon\); notice that for all \(t\) we get \[\begin{align*} \int\left\lvert\sum_{j=1}^k f(jx+ta)\right\rvert dx - \int\left\lvert \sum_{j=1}^k g(jx+ta)\right\rvert dx &\leq \int\left\lvert \sum_{j=1}^k f(jx+ta)-g(jx+ta)\right\rvert dx \\ &\leq \sum_{j=1}^k \int \left\lvert f(jx+ta)-g(jx+ta)\right\rvert dx \\ &= \sum_{j=1}^k \frac 1j \left\lVert f-g\right\rVert _{L^1} \\ &< \epsilon \sum_{j=1}^k \frac 1j. \end{align*}\] Now let \(R>0\) so that the support of \(g\) is contained within the ball of radius \(R\) (centred at the origin). Then, when \(t\) is large enough that \(\lVert at\rVert > (2k+1)R\), we get that \(jx+at\in\operatorname{supp}(g)\) implies \(\lVert jx\rVert > 2kR\) for all \(j=1,\ldots, k\). In particular, if \(jx+at\in\operatorname{supp}(g)\) for any \(j\), then \(\lVert x\rVert > 2R\), so none of the other \(jx+at\)’s are in the support of \(g\). Thus, when \(t\) is very large, we get that the sum contains at most one nonzero term for all \(x\), and hence \[\lim_{t\to\infty}\int\left\lvert \sum_{j=1}^k g(jx+at)\right\rvert dx = \lim_{t\to\infty} \sum_{j=1}^k \int\left\lvert g(jx+at)\right\rvert dx = \lVert g\rVert_{L^1} \sum_{j=1}^k \frac 1j.\] The limit for \(f\) is within \(\epsilon\sum_{j=1}^k\frac 1j\) of \(\lVert g\rVert_{L^1}\sum_{j=1}^k\frac 1j\), which is in turn within \(\epsilon\sum_{j=1}^k\frac 1j\) of \(\lVert f\rVert_{L^1}\sum_{j=1}^k\frac 1j\). Together, we get that \[\lim_{t\to\infty}\left\lvert\int\left\lvert\sum_{j=1}^k f(jx+at)\right\rvert dx - \lVert f\rVert_{L^1} \sum_{j=1}^k \frac 1j\right\rvert < 2\epsilon \sum_{j=1}^k \frac 1j.\] Sending \(\epsilon\to 0\) gives the limit.

**Problem 2. **

Let \(f\in L^p(\mathbb R)\) for some \(1\leq p<2\). Show that the series \[\sum_{n=1}^\infty \frac{f(x+n)}{\sqrt n}\] converges absolutely for almost all \(x\in\mathbb R\). For each \(2\leq p\leq\infty\), give an example of a function \(f\in L^p(\mathbb R)\) for which the series diverges for every \(x\in\mathbb R\).

This problem looked very intimidating to me at first glance, and I had a lot of trouble figuring out where to start. I ended up going to bed, and the solution came to me in a dry dream.

There are two things to notice:

- If \(q\) is conjugate to \(p\), then \(q>2\) and so \(\frac 1{\sqrt x}\) is in \(L^q\). (You’d have to truncate this function for \(q=\infty\)).
- If you can integrate the series on some subset of \(\mathbb R\), then you can deduce that the series converges for almost every \(x\) in that subset.

Together, these two realisations indicate that one should relate the sum to an integral, from where one may invoke Hӧlder’s inequality to justify convergence.

We’ll just show that the series converges absolutely on \([0, 1]\); the argument can be easily adapted to any unit-length interval (left as an exercise for the reader). Define the function \(g = \sum_{n=1}^\infty \frac 1{\sqrt n}\chi_{[n+1, n+2)}\). Notice that \(g\) is in \(L^q\), where \(2< q\leq\infty\) is conjugate to \(p\). For \(q=\infty\) this is obvious, and otherwise we get that \[\lVert g\rVert_q^q=\sum_{n=1}^\infty n^{-\frac q2}<\infty\] since \(\frac q2>1\). When we integrate, we get \[ \int_0^1\sum_{n=1}^\infty\left\lvert\frac{f(x+n)}{\sqrt n}\right\rvert dx = \sum_{n=1}^\infty \int_0^1 \lvert f(x+n)g(x+n)\rvert dx = \int_1^\infty \lvert f(x) g(x) \rvert dx.\] Note that we may swap the order of integration by Tonelli’s theorem. We now use Hӧlder’s inequality to see that \[\int_0^1 \sum_{n=1}^\infty \left\lvert\frac{f(x+n)}{\sqrt n}\right\rvert dx \leq \lVert f\rVert_p \lVert g\rVert_q < \infty,\] giving the claim.

For the second part of the problem, \(f=1\) works for \(p=\infty\); for \(2\leq p<\infty\), \(f(x)=x^{-\frac 1p+\epsilon}\) with a cutoff near \(x=0\) works. The series will resemble a diverging \(p\)-series for all \(x\). I’ll leave the details as an exercise.

**Problem 3. **

Fix \(K>0\) and let \(\mathcal M_K\left(\mathbb R^d\right)\) denote the space of finite positive Borel measures \(\mu\) on \(\mathbb R^d\) with \(\mu\left(\mathbb R^d\right)\leq K\). When \(\mu_1,\mu_2\in \mathcal M_K\left(\mathbb R^d\right)\), write \(\mu_1\leq\mu_2\) if \(\mu_1(U)\leq\mu_2(U)\) for all Borel \(U\subseteq\mathbb R^d\). Show that the set \[S=\left\lbrace \left(\mu_1,\mu_2\right)\in \mathcal M_K\left(\mathbb R^d\right)\times \mathcal M_K\left(\mathbb R^d\right) : \mu_1\leq \mu_2\right\rbrace\] is compact in the weak* topology. Here, \(\mathcal M_K\left(\mathbb R^d\right)\) is viewed as a subset of the space of finite Borel measures on \(\mathbb R^d\) under the weak* topology.

I have never in my life showed weak* compactness directly; I have only ever cited Alaoglu’s theorem or the uniform boundedness principle. In fact, I think these are the only two theorems that I know that are related to weak* compactness.

In this case, Alaoglu’s theorem is very handy, as it gives us compactness of \(\mathcal M_K\left(\mathbb R^d\right)\). In turn, we see that \(S\) is a subset of a compact space, and so it suffices to show that \(S\) is closed. Importantly, we must realise that the space of finite Borel measures on \(\mathbb R^d\) is the dual of the space of bounded continuous functions under the uniform norm. This function space is complete, and thus the conditions of the theorem are met.

Let \(\left(\mu_n, \nu_n\right)\) a sequence in \(S\) convering in weak* to some \((\mu, \nu)\in\mathcal M_K\left(\mathbb R^d\right)\times \mathcal M_K\left(\mathbb R^d\right)\). We wish to show that \(\mu\leq \nu\). But this is straightforward: by weak* convergence we get \(\lim \mu_n(U)=\mu(U)\) and \(\lim \nu_n(U) = \nu(U)\). But since \(\mu_n(U)\leq\nu_n(U)\), their limits obey the same inequality, and we conclude that \(\mu\leq\nu\). Thus \(S\) is a closed subset of a compact space, and hence it is compact itself.

**Problem 4. **

Suppose \(2\leq p<\infty\). If \(\mu,\nu\) are positive measures on \(\mathbb R^d\) and \(f, f_j\) are finitely many functions in \(L^p\left(\mathbb R^d\right)\) satisfying \(f=\sum_j f_j\), the inequality \[\lVert f\rVert_{L^p(\mu)} \leq M\left(\sum_j\lVert f_j\rVert_{L^p(\nu)}^2\right)^{\frac 12}\tag{1}\] is called an \(l^2L^p\) decoupling inequality and \(M>1\) is called the decoupling constant. Show that if \(\mu=\sum_k\mu_k\) and \(\nu=\sum_k\nu_k\), where the sums are finite, and the \(l^2L^p\) decoupling inequalities hold with decoupling constant \(M\), \[\lVert f\rVert_{L^p(\mu_k)} \leq M\left(\sum_j\lVert f_j\rVert_{L^p(\nu_k)}^2\right)^{\frac 12}\] for all \(k\), then (1) holds with the same decoupling constant \(M\).

To relate the two inequalities, we need to sum the \(L^p\left(\mu_k\right)\) norms together before getting an inequality involving the \(L^p(\mu)\) norm. The issue then becomes that the summations involving \(\lVert f_j\rVert_{L^p(\nu_k)}\) are in the wrong order and don’t mix well, and this requires some work to manipulate.

We begin with \[\lVert f\rVert_{L^p(\mu)}^p = \sum_k \lVert f\rVert_{L^p(\mu_k)}^p \leq M^p \sum_k \left(\sum_j \lVert f_j\rVert_{L^p(\nu_k)}^2\right)^{\frac p2}.\] We would like to swap the summations in \(k\) and \(j\) with each other, but the funny exponents make this troublesome. Consider instead the inequality \[\lVert f\rVert_{L^p(\mu)}^{\frac 2p}\leq M^{\frac 2p} \left(\sum_k\left(\sum_j g(j, k)\right)^{\frac p2}\right)^{\frac 2{p^2}},\] where \(g(j, k)=\lVert f_j\rVert_{L^p(\nu_k)}^2\). Then, treating the sums as integrals with respect to a counting measure on the sets of indices in \(j\) and \(k\), we may apply Minkowski’s inequality for integrals and get \[\lVert f\rVert_{L^p(\mu)}^{\frac 2p}\leq M^{\frac 2p}\left(\sum_j \left(\sum_k g(j, k)^{\frac p2}\right)^{\frac 2p}\right)^{\frac 1p}.\] Importantly, \(1\leq\frac p2<\infty\), so the inequality really is applicable. But the inner sum is just \(\sum_{k} \lVert f_j\rVert_{L^p(\nu_k)}^p=\lVert f_j\rVert_{L^p(\nu)}^p\), so the whole thing simplifies to \[\lVert f\rVert_{L^p(\mu}^{\frac 2p}\leq M^{\frac 2p}\left(\sum_j \lVert f_j\rVert_{L^p(\nu)}^2\right)^{\frac 1p},\] and (1) follows after raising both sides to the \(\frac p2\) power.

**Problem 5. **

Let us define \(Tf(x)=\int_0^x f(t) dt\) for \(x\in[0, 1]\) and \(f\in L^2([0, 1])\).

- Prove that \(T:L^2([0, 1])\to L^2([0, 1])\) is a linear continuous map which is compact, in the sense that for any bounded sequence \(f_n\in L^2([0, 1])\), the sequence \(Tf_n\) has a convergent subsequence in \(L^2([0, 1])\).
- Prove that \(T\) has no eigenvalues, i.e. prove that there is no \(\lambda\in\mathbb C\) such that \(Tf=\lambda f\) for some nonzero \(f\in L^2([0, 1])\).
- Show that the spectrum of \(T\) is \(\lbrace 0\rbrace\), i.e. show that the map \(f\mapsto Tf-\lambda f\) is an isomorphism of \(L^2([0, 1])\) if and only if \(\lambda\in\mathbb C\) is nonzero.

For part (a), it suffices to show that \(T\) is a bounded operator. Write \(Tf(x)=\int_{0}^{1}\chi_{t\leq x} f(t) dt\); then, we get by Minkowski’s integral inequality that \[\left\lVert Tf(x) \right\rVert_{L^2}\leq \int_{0}^{1}\left\lVert \chi_{x\geq t} f(t) \right\rVert_{L^2} dt.\] Clearly it follows that \(\left\lVert Tf \right\rVert_{L^2}\leq \left\lVert f \right\rVert_{L^2}\). Bounded linear operators on Banach spaces are continuous.

For compactness, let \(f_n\) be a sequence of functions in \(L^2([0, 1])\), and suppose \(\left\lVert f_n \right\rVert_{L^2}\leq M\) for some \(M>0\). We claim that \(Tf_n\) are uniformly bounded and equicontinuous; Arzela-Ascoli then gives a uniformly convergent subsequence, which implies \(L^2\) convergence since we’re on a compact domain.

For uniform boundedness, we have \[\left\lvert Tf_n(x) \right\rvert\leq \int_{0}^{x}\left\lvert f_n(x) \right\rvert dx \leq \int_{0}^{1}\left\lvert f_n(x) \right\rvert dx \leq \sqrt M\] by Cauchy-Schwarz. Similarly, for uniform equicontinuity, if \(x< y\) we get \[\left\lvert Tf_n(y)-Tf_n(x) \right\rvert\leq \int_{x}^{y} \left\lvert f_n(x) \right\rvert\leq \sqrt M \left\lvert x-y \right\rvert^{\frac{1}{2} }.\] This gives uniform equicontinuity, as desired. To show that \(T\) has no eigenvalues, suppose there is a solution to \(Tf=\lambda f\). If \(f\in L^2\), then \(Tf\) is continuous; if \(Tf\) is continuous, then \(f\) is continuous; if \(f\) is continuous, then \(Tf\) is continuously differentiable. By induction we see that \(f\) is smooth. But then \(f\) solves the ODE \(f’=\lambda f\), so it follows that \(f\) is a multiple of \(e^{\lambda x}\). Since \(Tf(0)=0\), it follows that \(f=0\).

Finally, for \(\lambda=0\) we get that \(Tf\) is continuous for all \(f\) and thus can’t be surjective — \(L^2\) is more than just continuous functions. Otherwise, one can directly and explicitly solve the equation \(Tf-\lambda f=g\) for \(g\in L^2\).

**Problem 6. **

Let \(E=\left\lbrace (x_1, x_2)\in\mathbb R^2 : x_1-x_2\notin\mathbb Q\right\rbrace\). Show that \(E\) does not contain a set of the form \(A_1\times A_2\), where \(A_1,A_2\subset\mathbb R\) are measurable and both of positive measure.

**Problem 7. **

Let \(\mathbb D=\left\lbrace z\in\mathbb C:\vert z\vert <1\right\rbrace\).

- Let \(f:\mathbb D\to\mathbb C\) be holomorphic injective and let \(g:\mathbb D\to\mathbb C\) be holomorphic such that \(g(0)=f(0)\) and \(g(\mathbb D)\subseteq f(\mathbb D)\). Show that \(g\left(\mathbb D_r\right)\subseteq f\left(\mathbb D_r\right)\) for all \(0< r< 1\), where \(\mathbb D_r=\lbrace z\in\mathbb C:\vert z\vert < r\rbrace\).
- Let \(g:\mathbb D\to\mathbb C\) be holomorphic such that \(g(0)=0\) and \(\vert\Re g(z)\vert<1\) for all \(z\in\mathbb D\). Show that \[\vert\Im g(z)\vert \leq \frac 2\pi \log\frac{1+\vert z\vert}{1-\vert z\vert}\] for all \(z\in\mathbb D\).

We have that \(f\) is a conformal map between \(\mathbb D\) and its image since \(f\) is injective. By applying \(f ^{-1}\), we have that \(f ^{-1} \circ f\) and \(f ^{-1} \circ g\) are both maps of the disc to itself, where here we use that the image of \(g\) is contained in the image of \(f\). Then, if \(z\in\mathbb D_r\), we have by the Schwarz lemma \[ \left\lvert f ^{-1}\circ g(z)\right \rvert \leq \vert z\vert\in \mathbb D_r=f ^{-1}\circ f\left(\mathbb D_r\right).\] Thus \(f ^{-1} \circ g\left(\mathbb D_r\right)\subseteq f ^{-1}\circ f\left(\mathbb D_r\right)\), and the first claim follows after applying \(f\) to both sides.

For the second claim, we would like to use part a; we thus need to find an injective function from \(\mathbb D\) to the strip \(\lbrace z\in\mathbb C : \vert\Re z\vert < 1\rbrace\). Fortunately, this can be sniffed out from the suggestive inequality.

In particular, we have \(z\mapsto\frac{1+z}{1-z}\) injectively maps the disk onto the right half-plane. Then, the standard branch of \(\operatorname{Log}\) maps the right half-plane to the strip \(\left\lbrace z\in\mathbb C:\vert\Re z\vert < \frac\pi 2\right\rbrace\). Finally, the factor of \(\frac 2\pi\) rescales it, and we get the map \(\varphi(z)=\frac 2\pi \operatorname{Log}\frac{1+z}{1-z}\) injectively maps \(\mathbb D\) onto the right strip.

Suppose \(z\in\mathbb D_r\). Then, we have that \[\vert\varphi(z)\vert = \frac 2\pi\log\frac{1+\vert z\vert}{1-\vert z\vert} < \frac 2\pi\log\frac{1+r}{1-r}\] for all \(\vert z\vert < r < 1\). Taking \(r\to\vert z\vert\), we get that \[\vert\varphi(z)\vert \leq \frac 2\pi \log\frac{1+\vert z\vert}{1-\vert z\vert}.\] Note that \(g(0)=\varphi(0)=0\), so by part a, the image of \(\mathbb D_r\) under \(g\) is contained within the image of \(\mathbb D_r\) under \(\varphi\), and thus the bounds in magnitude hold as well. This yields \[\vert\Im g(z)\vert \leq \vert g(z)\vert \leq \frac 2\pi\log\frac{1+\vert z\vert}{1-\vert z\vert},\] as desired.

**Problem 8. **

Show that \[\frac{\pi}{\sin(\pi z)}=\frac 1z+2z\sum_{n=1}^\infty\frac{(-1)^n}{z^2-n^2}\] for all \(z\in\mathbb C\setminus\mathbb Z\), with the series in the right hand side converging uniformly on compact subsets of \(\mathbb C\setminus\mathbb Z\).

**Problem 9. **

Let \(\Omega\subset\mathbb C\) be open connected and let \(f_j:\Omega\to\mathbb C\) be a sequence of holomorphic functions. Suppose that \(f_j(a)\) converges as \(j\to\infty\), for some \(a\in\Omega\), and that the sequence \(\Re f_j\) converges as \(j\to\infty\) uniformly on compact subsets of \(\Omega\). Show that \(f_j\) converges as \(j\to\infty\) uniformly on compact subsets of \(\Omega\).

**Problem 10. **

Let \(f:\mathbb C\to\mathbb C\) be entire and set \[m(r)=\frac 1{2\pi}\int_0^{2\pi}\log_+ \left\lvert f\left(re^{i\varphi}\right)\right\rvert d\varphi,\] where \(\log_+t=\max(\log t, 0)\). Suppose that \[\limsup_{r\to\infty}\frac{m(r)}{\log r} < \infty.\] Show that \(f\) is a polynomial.

**Problem 11. **

Let \(f\in C(\mathbb R)\cap L^1(\mathbb R)\), and define \[u(z)=\frac 1{2\pi i}\int_{-\infty}^\infty \frac{f(t)}{t-z} dt\] for \(\Im z\neq 0\).

- Prove that \(u\) is a holomorphic function on \(\mathbb C\setminus\mathbb R\) such that \(u(z)\to 0\) as \(\vert\Im z\vert\to\infty\).
- Show that the limit \[\lim_{y\to 0^+}\left(u(x+iy)-u(x-iy)\right)\] exists for each \(x\in\mathbb R\) and compute it.

For the first part, it suffices to show that the integral converges absolutely when \(\Im z\neq 0\). Well, \[\int_{-\infty}^\infty \left\lvert \frac{f(t)}{t-z}\right\rvert dt \leq \frac 1{\lvert\Im z\rvert}\int_{-\infty}^\infty \vert f(t)\vert dt,\] as the denominator is bounded in magnitude by \(\vert\Im z\vert\) from below. The remaining integral is just \(\Vert f\Vert_{L^1}<\infty\). Moreover, the same bound shows that \(u\to 0\) as \(\vert\Im z\vert\to\infty\).

For the limit, we may expand and collect the denominators inside the integral; we have \[\frac{1}{t-x-iy}-\frac{1}{t-x+iy}=\frac{2iy}{(t-x)^2+y^2}.\] Thus, after shifting by \(x\), we get that \[\lim_{y\to 0^+}\left(u(x+iy)-u(x-iy)\right)=\frac y\pi\int_{-\infty}^\infty\frac{f(t+x)}{t^2+y^2} dt.\] We claim that this approaches \(f(x)\). Notice that \(\int_{\mathbb{R}} \frac{1}{t^2+y^2} dt = \frac{\pi}{y}\); we’ll show that for all \(\epsilon>0\), we can find a \(y>0\) such that \[\left\lvert \frac{y}{\pi} \int_{-\infty}^{\infty}\frac{f(t+x)}{t^2+y^2}dt-f(x) \right\rvert= \left\lvert \frac{y}{\pi} \int _{-\infty}^{\infty} \frac{f(t+x)-f(x)}{t^2+y^2}dt \right\rvert<\epsilon.\] Since \(f\) is continuous at \(x\), there exists a \(\delta>0\) such that \(\left\lvert f(t+x)-f(x) \right\rvert<\epsilon\) for all \(\left\lvert t \right\rvert<\delta\). We now split the integral into three pieces: \[\int_{-\infty}^{\infty}\frac{f(t+x)-f(x)}{t^2+y^2}dt=\int_{-\delta}^{\delta}\frac{f(t+x)-f(x)}{t^2+y^2}dt+\int_{\left\lvert t \right\rvert>\delta}\frac{f(t+x)}{t^2+y^2}dt-\int_{\left\lvert t \right\rvert>\delta}\frac{f(x)}{t^2+y^2}dt.\] The first integral is bounded in magnitude by \(\frac{\epsilon\pi}{y} \): the \(\epsilon\) comes from the continuity, and the \(\frac{\pi}{y} \) comes from extending the integral and just integrating. The middle integral is bounded in magnitude by \(\delta^{-2}\left\lVert f \right\rVert_{L^1}\), as the denominator is always larger than \(\delta^2\). Finally, for the third integral, we may compute it explicitly with \(s=\frac{t}{y} \) as: \[\int_{\left\lvert t \right\rvert>\delta} \frac{1}{t^2+y^2} dt=\frac{1}{y} \int_{\left\lvert s \right\rvert>\frac{\delta}{y} } \frac{1}{s^2+1} ds=\frac{1}{y} \left( \pi-2\arctan\left( \frac{\delta}{y} \right) \right).\] Putting these all together, we get the inequality \[\left\lvert \frac{y}{\pi} \int_{-\infty}^{\infty}\frac{f(t+x)}{t^2+y^2}dt-f(x) \right\rvert<\epsilon+\frac{y \left\lVert f \right\rVert_{L^1}}{\delta^2\pi}+\pi f(x) \left( \pi-2\arctan\left( \frac{\delta}{y} \right) \right).\]

Take \(y\to 0\) to get the bound by \(\epsilon\). \(\epsilon>0\) was arbitrary, and so the limit really is \(f(x)\) as \(y\to 0\).

**Problem 12. **

Let \(f:\mathbb C\to\mathbb C\) be entire and assume that \(f\) is not of the form \(z\mapsto z+a\) for some \(a\in\mathbb C\).

- Show that the composition \(f\circ f\) has a fixed point.
- Find an entire function \(f:\mathbb C\to\mathbb C\) (not of the form \(z\mapsto z+a\) with no fixed points.

For part a, let \(f\) be a function such that \(f\circ f\) has no fixed points. Note that \(f\) can’t have fixed points — fixed points of \(f\) are fixed points of \(f\circ f\). Then, the function \(g(z)=\frac{f\circ f(z)-z}{f(z)-z}\) is entire and avoids zero, as neither sides of the fraction are ever zero. But \(g(z)=1\) implies \(f\circ f(z)=f(z)\), in turn implying \(f(z)\) is a fixed point of \(f\). Thus \(g\) avoids both \(0\) and \(1\), so \(g\) is constant by the little Picard theorem.

We can actually rewrite \(g(z) = \frac{f(f(z))-f(z)}{f(z)-z}+1\), so this fraction is still constant. Suppose \(f(z)=f(w)=a\) for some \(z, w\in \mathbb{C}\). Then, we get that \[\frac{f(a)-a}{a-z}=\frac{f(a)-a}{a-w}\implies z=w,\] so \(f\) is entire and injective. The only such functions are linear functions \(f=mz+c\). If \(m\neq 1\) or \(c=0\) then \(f\) would have a fixed point, so we conclude that \(f\) must be of the form \(f(z)=z+c\) for some nonzero \(c\in \mathbb{C}\).

Finally, for a function with no fixed points, we just need \(f(z)-z\) to be nonzero for all \(z\). Picking \(f(z)-z=e^z\), e.g. \(f(z)=z+e^z\), just works.